[tex]\bf \displaystyle \cfrac{d}{dx}\left[ \int\limits_{\sqrt{x}}^{\frac{\pi }{2}}~[cos(t^2)]dt \right]\implies \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{\sqrt{x}}~[cos(t^2)]dt \right]\\\\\\ \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{x^{\frac{1}{2}}}~[cos(t^2)]dt \right]\\\\
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u=x^{\frac{1}{2}}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}}\\\\\\ \textit{and now, let's use the \underline{2nd fundamental theorem of calculus}}\\\\
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[tex]\bf \displaystyle \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{u}~[cos(t^2)]dt \right]\impliedby \cfrac{df}{dx}=\cfrac{df}{du}\cdot \cfrac{du}{dx}
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\cfrac{df}{dx}=-cos(u^2)\cdot \cfrac{du}{dx}\implies \cfrac{df}{dx}=-cos(u^2)\cdot \cfrac{1}{2\sqrt{x}}
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\cfrac{df}{dx}=-cos[(\sqrt{x})^2]\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{df}{dx}=-\cfrac{cos(x)}{2\sqrt{x}}[/tex]
